HDU 6047 (2017 多校训练赛2 1003）Maximum Sequence

2017 Multi-University Training Contest - Team 2 1003

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .

Now Steph finds it too hard to solve the problem, please help him.

 

Input

The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

 

Output

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

 

Sample Input

4
8 11 8 5
3 1 4 2

 

Sample Output

27
Hint
For the first sample:
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

 

题意：

给出数列 A 和B

A代表着当前数列

B代表着从B[i]到整个A数列结束取最大的 (A[k]-k)添加到A的最后

问添加的数之和最大是多少

分析：

用样例来分析

a： 8 11 8 5

a[i]-i: 7  9  5  1

b：3 1 4 2

先将b排序得到  b: 1 2 3 4

那么取第一个b[0]=1 那么取到9

a：8 11 8 5 9

a[i]-i: 7 9 5 1 4

b[1]=2 还是取到 9

a: 8 11 8 5 9 9

a[i]-i: 7 9 5 1 4 3

后面继续这样取最终得到的是

a: 8 11 8 5 9 9 5 4  多出来的就是 27

那么从模拟的这个过程可以发现，只要记录当前位置到n中间的最大值 和更新出来的数中的最大值

进行比较取最大的那个，就是答案啦~

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define mod 1000000007 
using namespace std;
long long a[300000],b[300000];
long long maxa[300000];
int main()
{
	int n;
	while(scanf("%d",&n)==1)
	{
		memset(maxa,0,sizeof(maxa));
		for(int i=1;i<=n;i++)
		{
			long long aa;
			scanf("%lld",&aa);
			a[i]=aa-i;
		}
		for(int i=n;i>=1;i--)
		maxa[i]=max(maxa[i+1],a[i]);
		for(int i=1;i<=n;i++)
		{
			scanf("%lld",&b[i]);
		}
		sort(b+1,b+n+1);
		long long ans=0;
		ans+=maxa[b[1]];
		ans%=mod;
		long long t=maxa[b[1]]-n-1;
		for(int i=2;i<=n;i++)
		{
			 maxa[b[i]]=max(maxa[b[i]],t);
			 ans+=maxa[b[i]];
			 ans%=mod;
			 t=max(t,maxa[b[i]]-n-i);
		}
		printf("%lld\n",ans);
	}
}